Answer to the Question 03/02
SOAP BUBBLEThe question was:
Calculate the time of disappearance of a soap bubble connected with the atmosphere via a capillary.
(6/2002) The problem has been solved (13/5/02) by Alex Smolyanitskiy MS student in in Electrical Engineering, Columbia University (NY) (e-mail shurakbh@hotmail.com) (the solution below closely follows his solution), and (22/6/02) by Chetan Mandayam Nayakar (e-mail mn_chetan@yahoo.com).
The solution: In this solution we assume that the process is sufficiently slow that we can
(a) disregard the effects of viscosity of the soap bubble and the times required for liquid in the wall of the bubble to readjust its shape.
(b) assume that the process is almost static despite constantly changing volume of the bubble. E.g., below we use Laplace equation and assume that the pressure is constant everywhere inside the bubble.
(c) assume that the air flow in the capillary is laminary.
The pressure difference between inside and outside of a spherical bubble is given by the Laplace equation p=4s/r,
where s is the surface tension of soapy water (approximately 0.025N/m), r is the radius of the bubble. In the above equation we introduced an additional prefactor 2 to account for the fact that the bubble has two surfaces. If we disregard the viscosity of the air and assume laminar flow, Bernoulli relation can be used to relate this pressure difference with the flow velocity v inside the capillary.
p=D·v2/2,
where D is the density of the air (1.3 kg/m3). If the cross section area of the capillary is A then we can easily relate the outflow of air via capillary with the decrease in the radius of the bubble
v·A·dt=-4{pi}·r2·dr
The above three equations can be simply combined into a single differential equation, which determines the rate of change of the radius of the bubble:
r5/2·dr=-(A/{pi})·(s/2D)1/2dt
This equation can be directly integrated by assuming that during time T the radius of the bubble changes from its initial value R to zero, leading to
T=(2{pi}/7A)·(2D·R/s)1/2·R3
As an example, let us consider a bubble of diameter R=0.05m, connected to a capillary of 1mm radius, i.e. A=3·10-6 m2. By substituting all those numbers into the above result we find T=80 sec.
Alex Smolyanitskiy experimentally verified the above theoretical expression for several values of soap bubble diameters.
This result has been obtained by assuming that air flow inside the capillary has everywhere the same velocity v. A more careful analysis will show the velocity does depend on the distance from the central axis of the capillary. This effect can be effectively taken into account by using smaller (effective) cross section of the capillary.
One can directly estimate the Reynolds number for typical flow velocities predicted by this treatment, and find the for buble radius ranging between 1 and 10 cm, and capillary diameters of few milimeters, the Reynolds number will be between 1 and 10, i.e. low enough to ensure laminar flow that was assumed in this derivation.
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